A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence point, 64.78mL of base were added. What is the molarity of the base?

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**Titration Problem: Determining the Molarity of a Base**

A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence point, 64.78 mL of base were added. What is the molarity of the base?

**[Answer Box]**

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This question involves a titration process where sulfuric acid (H₂SO₄) is neutralized by potassium hydroxide (KOH). To solve this, consider the reaction:

\[ H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O \]

From this balanced equation, it is evident that one mole of sulfuric acid reacts with two moles of potassium hydroxide.

Given:
- Volume of H₂SO₄ = 25.00 mL = 0.02500 L
- Molarity of H₂SO₄ = 1.637 M
- Volume of KOH at equivalence point = 64.78 mL = 0.06478 L

To find the molarity of KOH, use the formula:

\[ M_1V_1 = M_2V_2 \]
\[ (1.637 \, \text{M})(0.02500 \, \text{L}) = M_{\text{KOH}}(0.06478 \, \text{L}) \]

\[ M_{\text{KOH}} = \frac{(1.637 \, \text{M})(0.02500 \, \text{L})}{(0.06478 \, \text{L})} = ? \]

Solve for the molarity of the potassium hydroxide solution.
Transcribed Image Text:**Titration Problem: Determining the Molarity of a Base** A 25.00 mL volume of 1.637 M sulfuric acid is titrated with potassium hydroxide. At the equivalence point, 64.78 mL of base were added. What is the molarity of the base? **[Answer Box]** [Include a space here for the answer box to be placed when implemented on the website] This question involves a titration process where sulfuric acid (H₂SO₄) is neutralized by potassium hydroxide (KOH). To solve this, consider the reaction: \[ H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O \] From this balanced equation, it is evident that one mole of sulfuric acid reacts with two moles of potassium hydroxide. Given: - Volume of H₂SO₄ = 25.00 mL = 0.02500 L - Molarity of H₂SO₄ = 1.637 M - Volume of KOH at equivalence point = 64.78 mL = 0.06478 L To find the molarity of KOH, use the formula: \[ M_1V_1 = M_2V_2 \] \[ (1.637 \, \text{M})(0.02500 \, \text{L}) = M_{\text{KOH}}(0.06478 \, \text{L}) \] \[ M_{\text{KOH}} = \frac{(1.637 \, \text{M})(0.02500 \, \text{L})}{(0.06478 \, \text{L})} = ? \] Solve for the molarity of the potassium hydroxide solution.
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